2012 AMC 10A Problem 22

1. The two sums. Look at small cases. The odd numbers: $1 = 1$, $1 + 3 = 4$, $1 + 3 + 5 = 9$. The first $m$ odd numbers add to $m^2$. The even numbers: $2 = 1 \cdot 2$, $2 + 4 = 6 = 2 \cdot 3$, $2 + 4 + 6 = 12 = 3 \cdot 4$. The first $n$ even numbers add to $n(n + 1)$, because $2 + 4 + \cdots + 2n$ is twice $1 + 2 + \cdots + n$.
2. The equation. $m^2 = n^2 + n + 212$.
3. Make a square. The right side is almost a perfect square. Multiply everything by $4$, because $4n^2 + 4n + 1 = (2n + 1)^2$:
$$4m^2 = 4n^2 + 4n + 848 = (2n + 1)^2 + 847.$$
So $(2m)^2 - (2n + 1)^2 = 847$.
4. Factor. A difference of two squares factors: $(2m - 2n - 1)(2m + 2n + 1) = 847$.
5. Split 847. $847 = 7 \times 121 = 7 \times 11 \times 11$. The ways to write it as small factor times large factor are
$$1 \times 847, \qquad 7 \times 121, \qquad 11 \times 77.$$
The second bracket is the larger one, so call the small factor $S = 2m - 2n - 1$ and the large one $L = 2m + 2n + 1$.
6. Solve each case. Subtracting, $L - S = 4n + 2$, so $n = \frac{L - S - 2}{4}$. Adding, $L + S = 4m$, so $m = \frac{L + S}{4}$.

1. All three give positive whole numbers. The possible values of $n$ add to $211 + 28 + 16 = 255$.

The answer is $\boxed{\textbf{(A)}\ 255}$.

When a whole-number equation has two squares that are close together, turn it into "square minus square equals a number". Multiplying by $4$ avoids fractions when completing the square. A difference of squares factors into two brackets, and a fixed number has only a few factor pairs, so the endless search becomes a short list.

Let $k = m - n$, which is at least $1$. Then $m^2 - n^2 = (m - n)(m + n) = k(2n + k)$, and this must equal $n + 212$. Rearranging, $n(2k - 1) = 212 - k^2$, so
$$n = \frac{212 - k^2}{2k - 1}.$$
Try $k = 1, 2, \dots, 14$ (for $k = 15$ the top is negative). Only $k = 1, 4, 6$ give whole numbers: $n = 211$, $\frac{196}{7} = 28$ and $\frac{176}{11} = 16$. The sum is again $255$.